Question

In problem 7-16, solve the equation. $\mathbf{x}\frac{\mathrm{dv}}{\mathrm{dx}}\mathbf{=}\frac{\mathbf{1}\mathbf{-}\mathbf{4}{\mathbf{v}}^2}{3v}$

Short answer

The solution of the given differential equation is$\mathbf{v}\mathbf{=}\mathbf{\pm }\frac{\sqrt{\mathbf{1}\mathbf{-}{\mathbf{Cx}}^{\mathbf{-}\frac{8}{3}}}}{2}$.

Step-by-step solution

Concept of Separable Differential Equation

A first-order ordinary differential equation$\frac{\mathrm{dy}}{\mathrm{dx}}\mathbf{=}\mathbf{f}\left(x,y\right)$ is referred to as separable if the function in the right-hand side of the equation is expressed as a product of two functions$\mathbf{g}\left(x\right)$ that is a function of x alone and$\mathbf{h}\left(y\right)$ that is a function of y alone.

A separable differential equation can be expressed as $\frac{\mathrm{dy}}{\mathrm{dx}}\mathbf{=}\mathbf{g}\left(x\right)\mathbf{h}\left(y\right)$. By separating the variables, the equation follows. Then, on direct integration of both sides, the solution of the differential equation is determined.

Solution of the Equation

The given equation is

$\mathrm{x}\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1-4{\mathrm{v}}^2}{3\mathrm{v}}\cdot \cdot \cdot \cdot \cdot \cdot \left(1\right)$

After separating the variables, equation (1) can be written as

$\frac{3\mathrm{v}}{1-4{\mathrm{v}}^2}\mathrm{dv}=\frac{\mathrm{dx}}{\mathrm{x}}\cdot \cdot \cdot \cdot \cdot \cdot \left(2\right)$

Integrate both sides of equation (2). It results,

$\begin{array}{c}\int \frac{3\mathrm{v}}{1-4{\mathrm{v}}^2}\mathrm{dv}=\int \frac{dx}{x}\\ -\frac{3}{8}\int \frac{-8\mathrm{v}\mathrm{dv}}{1-4{\mathrm{v}}^2}=\int \frac{dx}{x}\\ -\frac{3}{8}\int \frac{\mathrm{d}\left(1-4{\mathrm{v}}^2\right)}{1-4{\mathrm{v}}^2}=\int \frac{dx}{x}\\ -\frac{3}{8}\ln \left(1-4{\mathrm{v}}^2\right)=\ln \mathrm{x}+\ln \mathrm{k}\left[\ln \mathrm{k}=\mathrm{IntegratingConstant}\right]\end{array}$

$\begin{array}{c}\ln \left(1-4{\mathrm{v}}^2\right)=-\frac{8}{3}\ln \mathrm{x}+\left(-\frac{8}{3}\right)\ln \mathrm{k}\\ \ln \left(1-4{\mathrm{v}}^2\right)=\ln {\mathrm{x}}^{-\frac{8}{3}}+\ln \mathrm{C}\left[\ln \mathrm{C}=-\frac{8}{3}\ln \mathrm{k}=\mathrm{Constant}\right]\\ \ln \left(1-4{\mathrm{v}}^2\right)-\ln \mathrm{C}=\ln {\mathrm{x}}^{-\frac{8}{3}}\\ \ln \left(\frac{1-4{\mathrm{v}}^2}{\mathrm{C}}\right)=\ln {\mathrm{x}}^{-\frac{8}{3}}\end{array}$

$\begin{array}{c}\frac{1-4{\mathrm{v}}^2}{C}={\mathrm{x}}^{-\frac{8}{3}}\\ 4{\mathrm{v}}^2=1-\mathrm{C}{\mathrm{x}}^{-\frac{8}{3}}\\ \mathrm{v}=\pm \frac{\sqrt{1-{\mathrm{Cx}}^{-\frac{8}{3}}}}{2}\end{array}$

Therefore, the solution of the given equation is$\mathbf{v}\mathbf{=}\mathbf{\pm }\frac{\sqrt{\mathbf{1}\mathbf{-}{\mathbf{Cx}}^{\mathbf{-}\frac{8}{3}}}}{2}$.